Cell Biology — Diagnostic Tests

Unit Tests

UT-1: Cell Structure and Microscopy

Question:

(a) Describe the structure and function of the following organelles: nucleus, mitochondrion, ribosome, and cell membrane.

(b) Explain the differences between light microscopy and electron microscopy, including resolution, magnification, specimen preparation, and the type of image produced.

(c) Calculate the actual size of a cell from a microscope image given the following information: the image of the cell measures $4.2\,\text{cm}$ on a printed micrograph, and the magnification is $\times 5000$. Give your answer in micrometres ($\mu\text{m}$).

(d) Describe the key features of a prokaryotic cell that distinguish it from a eukaryotic cell. Give at least four differences.

Solution:

(a)

• Nucleus: A large, membrane-bound organelle containing the cell’s genetic material (DNA) organised into chromosomes. It is surrounded by a double membrane (nuclear envelope) with nuclear pores that control the movement of substances in and out. The nucleolus within the nucleus is the site of ribosomal RNA synthesis. Function: controls cell activities by regulating gene expression and stores the genetic information.
• Mitochondrion: A double-membrane organelle; the inner membrane is folded into cristae to increase surface area for the reactions of aerobic respiration. The matrix contains enzymes for the Krebs cycle, mitochondrial DNA, and ribosomes. Function: produces ATP through aerobic respiration — the “powerhouse” of the cell.
• Ribosome: A small organelle composed of rRNA and protein, found free in the cytoplasm or attached to the rough ER. It consists of a large and small subunit. Function: site of protein synthesis (translation), where mRNA is read and amino acids are assembled into polypeptide chains.
• Cell membrane: A phospholipid bilayer with embedded proteins (fluid mosaic model). It is selectively permeable. Function: controls the movement of substances in and out of the cell, provides a barrier, and facilitates cell recognition and signalling.

(b)

Feature	Light Microscope	Electron Microscope
Resolution	Up to $0.2\,\mu\text{m}$ (200 nm)	Up to $0.2\,\text{nm}$ (TEM)
Magnification	Up to $\times 1500$	Up to $\times 2{,}000{,}000$
Specimen	Living or dead; relatively simple preparation	Dead only; complex preparation (fixation, dehydration, staining, embedding in resin)
Image	Coloured (natural or stained); image seen directly	Black and white (TEM) or 3D surface (SEM); image viewed on fluorescent screen or monitor
Vacuum	Not required	Required (electron beam travels in vacuum)
Radiation	Light (photons)	Beam of electrons

(c) Image size $= 4.2\,\text{cm} = 42{,}000\,\mu\text{m}$.

$\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}} = \frac{42000}{5000} = 8.4\,\mu\text{m}$

(d) Four differences between prokaryotic and eukaryotic cells:

1. Nucleus: Eukaryotic cells have a membrane-bound nucleus; prokaryotic cells do not (DNA is in a nucleoid region).
2. DNA structure: Eukaryotic DNA is linear and associated with histones as chromosomes; prokaryotic DNA is circular and not associated with histones. Prokaryotes may also have plasmids (small extra circular DNA).
3. Membrane-bound organelles: Eukaryotic cells have mitochondria, ER, Golgi, and other organelles; prokaryotic cells have none.
4. Ribosomes: Eukaryotic ribosomes are 80S (larger); prokaryotic ribosomes are 70S (smaller).
5. Cell wall: If present, eukaryotic cell walls are made of cellulose (plants) or chitin (fungi); prokaryotic cell walls are made of peptidoglycan.
6. Size: Prokaryotic cells are typically $0.5$—$5\,\mu\text{m}$; eukaryotic cells are typically $10$—$100\,\mu\text{m}$.

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UT-2: Biological Molecules

Question:

(a) Describe the structure of a generalised amino acid. Explain how a peptide bond is formed during protein synthesis.

(b) Describe the structure of a triglyceride and explain the difference between a saturated and an unsaturated fatty acid.

(c) Describe the test for reducing sugars (Benedict’s test) and the test for starch (iodine test). State the expected results for a positive test in each case.

(d) Explain the role of hydrogen bonding in determining the secondary and tertiary structure of proteins.

Solution:

(a) A generalised amino acid has a central carbon atom (the alpha carbon) bonded to four groups: an amino group ($\text{NH}_2$), a carboxyl group ($\text{COOH}$), a hydrogen atom, and a variable R group (side chain) that differs between the 20 amino acids. During protein synthesis, a peptide bond is formed by a condensation reaction between the amino group of one amino acid and the carboxyl group of another. A molecule of water ($\text{H}_2\text{O}$) is released. The resulting bond is a covalent $\text{C}-\text{N}$ bond linking the amino acids: $-\text{CO}-\text{NH}-$.

(b) A triglyceride consists of one glycerol molecule esterified to three fatty acid molecules. Glycerol is a 3-carbon alcohol with hydroxyl groups on each carbon. Each fatty acid has a carboxyl group that bonds to a hydroxyl on glycerol via a condensation reaction, releasing water.

Saturated fatty acids have no double bonds between carbon atoms in the hydrocarbon chain; all carbon atoms are bonded to the maximum number of hydrogen atoms. The chain is straight, allowing triglycerides to pack closely, making them solid at room temperature (e.g., butter). Unsaturated fatty acids contain one or more $\text{C}=\text{C}$ double bonds, creating kinks in the chain that prevent tight packing, making them liquid at room temperature (e.g., olive oil).

(c) Benedict’s test for reducing sugars: Add Benedict’s reagent (blue) to the sample and heat in a water bath for 5 minutes. A positive test produces a colour change from blue through green, yellow, and orange to a brick-red precipitate. The colour indicates the concentration of reducing sugar present.

Iodine test for starch: Add iodine solution (brown-orange) to the sample. A positive test produces a blue-black colour change, indicating the presence of starch.

(d) Hydrogen bonds form between the oxygen of one $\text{C}=\text{O}$ group and the hydrogen of an $\text{N}-\text{H}$ group on different parts of the polypeptide chain. In secondary structure, hydrogen bonds between the backbone atoms form regular patterns: alpha helices (spiral structure held by hydrogen bonds parallel to the helix axis) and beta pleated sheets (hydrogen bonds between adjacent strands). In tertiary structure, hydrogen bonds form between R groups and the backbone, helping fold the polypeptide into its specific 3D shape. Hydrogen bonds are relatively weak individually but collectively contribute significantly to the stability of protein structure. Other bonds involved include ionic bonds, disulphide bridges, and hydrophobic interactions.

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UT-3: Cell Membrane and Transport

Question:

(a) Describe the fluid mosaic model of the cell membrane. Name the main components and explain the roles of cholesterol and intrinsic (transmembrane) proteins.

(b) Define the following terms: diffusion, osmosis, active transport, facilitated diffusion.

(c) Plant cells have a cell wall in addition to the cell membrane. Explain the terms “turgid,” “flaccid,” and “plasmolysed” in the context of plant cells in solutions of different water potentials.

(d) Explain why the cell membrane is described as “selectively permeable” rather than “fully permeable” or “impermeable.”

Solution:

(a) The fluid mosaic model describes the cell membrane as a bilayer of phospholipids with proteins embedded in it. The main components are:

• Phospholipids: form the bilayer, with hydrophilic heads facing outward and hydrophobic tails facing inward. They can move laterally within the bilayer (fluid nature).
• Cholesterol: positioned between phospholipids, modulating membrane fluidity. At low temperatures, it prevents the phospholipids from packing too tightly and becoming too rigid. At high temperatures, it restricts excessive movement and maintains stability.
• Intrinsic (transmembrane) proteins: span the entire bilayer and function as channels or carriers for transporting molecules across the membrane, as receptors for cell signalling, and as enzymes. They are embedded within the “mosaic” of the membrane.

(b)

• Diffusion: the net movement of molecules from a region of higher concentration to a region of lower concentration, down the concentration gradient. Requires no energy (passive process).
• Osmosis: the net movement of water molecules across a selectively permeable membrane from a region of higher water potential to a region of lower water potential.
• Active transport: the movement of molecules or ions across a membrane against the concentration gradient, from lower to higher concentration. Requires energy in the form of ATP and carrier proteins.
• Facilitated diffusion: the passive movement of molecules across a membrane via specific transport proteins (channel or carrier proteins), down the concentration gradient. Does not require ATP.

(c)

• Turgid: when a plant cell is placed in a hypotonic solution (high water potential), water enters by osmosis. The vacuole swells and pushes the cytoplasm against the cell wall. The cell wall exerts an inward pressure (wall pressure) that prevents further water entry. The cell is firm and rigid, which is important for plant support.
• Flaccid: when a plant cell is in an isotonic solution, there is no net movement of water. The cell is not swollen or shrunk; it lacks turgor pressure and is limp.
• Plasmolysed: when a plant cell is placed in a hypertonic solution (low water potential), water leaves the cell by osmosis. The vacuole shrinks, the cell membrane pulls away from the cell wall, and the cytoplasm becomes concentrated. The cell is plasmolysed.

(d) The cell membrane is selectively permeable because it allows some substances to pass through freely (small non-polar molecules like $\text{O}_2$ and $\text{CO}_2$), allows others to pass under specific conditions (via transport proteins), and completely blocks others (large molecules and ions cannot pass through the hydrophobic core without a transport protein). It is not fully permeable (which would allow everything through) or impermeable (which would allow nothing through). The selective permeability is essential for maintaining the internal environment of the cell.

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Integration Tests

IT-1: Biological Molecules and Cell Function

Question:

(a) Enzymes are proteins that catalyse biological reactions. Explain how the tertiary structure of an enzyme is essential for its function, and describe what happens when an enzyme is denatured.

(b) A student tests four solutions for the presence of biological molecules. Solution A gives a blue-black result with iodine. Solution B gives a brick-red precipitate with Benedict’s reagent. Solution C turns biuret reagent purple. Solution D gives no positive result with any test. Identify the biological molecule present (if any) in each solution.

(c) The phospholipid bilayer of the cell membrane allows oxygen to pass through freely but not glucose. Explain this difference in terms of the molecular properties of each substance.

(d) A biologist wants to determine whether a sample of tissue is from a plant or an animal source. Describe three structural features they could observe under a microscope that would distinguish the two.

Solution:

(a) The tertiary structure of an enzyme determines the precise 3D shape of its active site, which is complementary to the shape of its specific substrate (lock and key model). The active site’s shape is maintained by hydrogen bonds, ionic bonds, disulphide bridges, and hydrophobic interactions. When an enzyme is denatured (by high temperature, extreme pH, or heavy metals), these bonds break, the tertiary structure unravels, and the active site changes shape. The substrate can no longer fit, and the enzyme loses its catalytic function. Denaturation is typically irreversible.

(b)

• Solution A: Starch (positive iodine test, blue-black).
• Solution B: Reducing sugar such as glucose or maltose (positive Benedict’s test, brick-red precipitate).
• Solution C: Protein (positive biuret test, purple — biuret reagent contains copper sulphate and sodium hydroxide, which react with peptide bonds).
• Solution D: No detectable carbohydrate, protein, or lipid under these tests. May contain non-reducing sugar (which would need hydrolysis first) or a substance not tested for (e.g., lipid).

(c) Oxygen ($\text{O}_2$) is a small, non-polar molecule that can diffuse directly through the hydrophobic interior of the phospholipid bilayer. Glucose is a larger, polar molecule that cannot pass through the hydrophobic core of the bilayer. Glucose requires a transport protein (carrier protein) for facilitated diffusion or active transport to cross the membrane.

(d) Three distinguishing features:

1. Cell wall: Plant cells have a rigid cell wall made of cellulose, visible as a distinct outer layer. Animal cells have no cell wall, only a cell membrane.
2. Chloroplasts: Plant cells (especially in leaf tissue) contain chloroplasts with visible grana. Animal cells never contain chloroplasts.
3. Central vacuole: Mature plant cells typically have a large central vacuole that occupies most of the cell volume, pushing the cytoplasm to the edges. Animal cells have small, multiple vacuoles (if any).
4. Shape: Plant cells are typically rectangular or angular due to the rigid cell wall. Animal cells are typically irregular or rounded.

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IT-2: Experimental Techniques in Cell Biology

Question:

(a) Describe how you would prepare a temporary wet mount of onion epidermal tissue for observation under a light microscope. Include the steps from obtaining the tissue to viewing the slide.

(b) A student observes red blood cells in three solutions under a microscope. In solution X, the cells appear normal. In solution Y, the cells appear swollen and some have burst. In solution Z, the cells appear shrunken and crenated. Identify the relative water potentials of solutions X, Y, and Z.

(c) Explain why samples for transmission electron microscopy (TEM) must be very thin. Describe how ultramicrotomy achieves this.

(d) A student wants to determine the effect of temperature on the rate of reaction of the enzyme amylase on starch. Outline the method, including how they would measure the rate and what controls they would use.

Solution:

(a) Method for preparing a temporary wet mount of onion epidermal tissue:

1. Peel a thin layer of epidermal tissue from the inner surface of an onion scale.
2. Place the tissue on a clean glass microscope slide.
3. Add one or two drops of iodine solution (to stain the cells and highlight structures).
4. Carefully lower a coverslip at an angle to avoid trapping air bubbles.
5. Blot excess liquid from around the coverslip with filter paper.
6. Place the slide on the microscope stage and start with the lowest power objective lens.
7. Focus using the coarse adjustment, then fine adjustment.
8. Increase magnification as needed to observe cell structures clearly.

(b)

• Solution X: Isotonic (same water potential as the cell cytoplasm) — no net movement of water, cells appear normal.
• Solution Y: Hypotonic (higher water potential than the cells) — water enters by osmosis, cells swell and may burst (lyse), since animal cells lack a rigid cell wall.
• Solution Z: Hypertonic (lower water potential than the cells) — water leaves by osmosis, cells shrink and become crenated.

(c) TEM uses a beam of electrons that can only penetrate very thin specimens. If the sample is too thick, electrons are scattered or absorbed unevenly, producing a blurry image. Ultramicrotomy involves embedding the fixed tissue in a hard resin block and cutting ultra-thin sections (typically $50$—$100\,\text{nm}$ thick) using a glass or diamond knife mounted on an ultramicrotome. The sections are then floated onto a copper grid for viewing.

(d) Method:

1. Prepare a series of water baths at different temperatures (e.g., $10, 20, 30, 40, 50, 60\,^\circ\text{C}$).
2. Prepare a starch solution and an amylase solution.
3. Add a fixed volume of starch solution to a test tube and a fixed volume of amylase to a separate test tube. Place both in the same water bath for 5 minutes to equilibrate.
4. Mix the solutions and start a timer.
5. At regular intervals (e.g., every 30 seconds), remove a drop of the mixture and test with iodine solution on a spotting tile.
6. Record the time taken for the iodine to stop turning blue-black (indicating all starch has been hydrolysed).
7. Repeat each temperature three times for reliability.
8. Calculate the rate as $1/\text{time}$ (the reciprocal of the time for starch to disappear).

Controls: use a tube with boiled amylase (denatured enzyme) to confirm that starch breakdown is enzymatic, not spontaneous. Keep all variables except temperature constant (concentration of amylase and starch, volume, pH).

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

• Confusing magnification with resolution: magnification makes an image larger; resolution determines the ability to distinguish between two close points.
• Forgetting that osmosis refers only to water movement, not solute movement.
• Stating that enzymes are “killed” when denatured — enzymes are not alive; they lose function because their active site changes shape.
• Describing plant cells as “bursting” in hypotonic solutions — plant cells become turgid due to the cell wall; only animal cells burst.
• Confusing the biuret test (for protein) with Benedict’s test (for reducing sugars).